Integrand size = 33, antiderivative size = 161 \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^2(c+d x)}{(a+a \cos (c+d x))^4} \, dx=-\frac {4 A \text {arctanh}(\sin (c+d x))}{a^4 d}+\frac {2 (332 A+3 C) \tan (c+d x)}{105 a^4 d}-\frac {(88 A-3 C) \tan (c+d x)}{105 a^4 d (1+\cos (c+d x))^2}-\frac {4 A \tan (c+d x)}{a^4 d (1+\cos (c+d x))}-\frac {(A+C) \tan (c+d x)}{7 d (a+a \cos (c+d x))^4}-\frac {2 (6 A-C) \tan (c+d x)}{35 a d (a+a \cos (c+d x))^3} \] Output:
-4*A*arctanh(sin(d*x+c))/a^4/d+2/105*(332*A+3*C)*tan(d*x+c)/a^4/d-1/105*(8 8*A-3*C)*tan(d*x+c)/a^4/d/(1+cos(d*x+c))^2-4*A*tan(d*x+c)/a^4/d/(1+cos(d*x +c))-1/7*(A+C)*tan(d*x+c)/d/(a+a*cos(d*x+c))^4-2/35*(6*A-C)*tan(d*x+c)/a/d /(a+a*cos(d*x+c))^3
Leaf count is larger than twice the leaf count of optimal. \(680\) vs. \(2(161)=322\).
Time = 8.70 (sec) , antiderivative size = 680, normalized size of antiderivative = 4.22 \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^2(c+d x)}{(a+a \cos (c+d x))^4} \, dx =\text {Too large to display} \] Input:
Integrate[((A + C*Cos[c + d*x]^2)*Sec[c + d*x]^2)/(a + a*Cos[c + d*x])^4,x ]
Output:
((128*A*Cos[c/2 + (d*x)/2]^8*Cos[c + d*x]^2*Log[Cos[c/2 + (d*x)/2] - Sin[c /2 + (d*x)/2]]*(C + A*Sec[c + d*x]^2))/(d*(1 + Cos[c + d*x])^4*(2*A + C + C*Cos[2*c + 2*d*x])) - (128*A*Cos[c/2 + (d*x)/2]^8*Cos[c + d*x]^2*Log[Cos[ c/2 + (d*x)/2] + Sin[c/2 + (d*x)/2]]*(C + A*Sec[c + d*x]^2))/(d*(1 + Cos[c + d*x])^4*(2*A + C + C*Cos[2*c + 2*d*x])) + (Cos[c/2 + (d*x)/2]*Cos[c + d *x]*Sec[c/2]*Sec[c]*(C + A*Sec[c + d*x]^2)*(-10780*A*Sin[(d*x)/2] - 210*C* Sin[(d*x)/2] + 18788*A*Sin[(3*d*x)/2] + 252*C*Sin[(3*d*x)/2] - 20524*A*Sin [c - (d*x)/2] - 126*C*Sin[c - (d*x)/2] + 14644*A*Sin[c + (d*x)/2] + 126*C* Sin[c + (d*x)/2] - 16660*A*Sin[2*c + (d*x)/2] - 210*C*Sin[2*c + (d*x)/2] - 4690*A*Sin[c + (3*d*x)/2] + 14378*A*Sin[2*c + (3*d*x)/2] + 252*C*Sin[2*c + (3*d*x)/2] - 9100*A*Sin[3*c + (3*d*x)/2] + 11668*A*Sin[c + (5*d*x)/2] + 132*C*Sin[c + (5*d*x)/2] - 630*A*Sin[2*c + (5*d*x)/2] + 9358*A*Sin[3*c + ( 5*d*x)/2] + 132*C*Sin[3*c + (5*d*x)/2] - 2940*A*Sin[4*c + (5*d*x)/2] + 422 8*A*Sin[2*c + (7*d*x)/2] + 42*C*Sin[2*c + (7*d*x)/2] + 315*A*Sin[3*c + (7* d*x)/2] + 3493*A*Sin[4*c + (7*d*x)/2] + 42*C*Sin[4*c + (7*d*x)/2] - 420*A* Sin[5*c + (7*d*x)/2] + 664*A*Sin[3*c + (9*d*x)/2] + 6*C*Sin[3*c + (9*d*x)/ 2] + 105*A*Sin[4*c + (9*d*x)/2] + 559*A*Sin[5*c + (9*d*x)/2] + 6*C*Sin[5*c + (9*d*x)/2]))/(840*d*(1 + Cos[c + d*x])^4*(2*A + C + C*Cos[2*c + 2*d*x]) ))/a^4
Time = 1.40 (sec) , antiderivative size = 184, normalized size of antiderivative = 1.14, number of steps used = 16, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.455, Rules used = {3042, 3521, 3042, 3457, 3042, 3457, 27, 3042, 3457, 3042, 3227, 3042, 4254, 24, 4257}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
| \(\displaystyle \int \frac {\sec ^2(c+d x) \left (A+C \cos ^2(c+d x)\right )}{(a \cos (c+d x)+a)^4} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {A+C \sin \left (c+d x+\frac {\pi }{2}\right )^2}{\sin \left (c+d x+\frac {\pi }{2}\right )^2 \left (a \sin \left (c+d x+\frac {\pi }{2}\right )+a\right )^4}dx\) |
| \(\Big \downarrow \) 3521 |
| \(\displaystyle \frac {\int \frac {(a (8 A+C)-a (4 A-3 C) \cos (c+d x)) \sec ^2(c+d x)}{(\cos (c+d x) a+a)^3}dx}{7 a^2}-\frac {(A+C) \tan (c+d x)}{7 d (a \cos (c+d x)+a)^4}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {a (8 A+C)-a (4 A-3 C) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^2 \left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^3}dx}{7 a^2}-\frac {(A+C) \tan (c+d x)}{7 d (a \cos (c+d x)+a)^4}\) |
| \(\Big \downarrow \) 3457 |
| \(\displaystyle \frac {\frac {\int \frac {\left (a^2 (52 A+3 C)-6 a^2 (6 A-C) \cos (c+d x)\right ) \sec ^2(c+d x)}{(\cos (c+d x) a+a)^2}dx}{5 a^2}-\frac {2 a (6 A-C) \tan (c+d x)}{5 d (a \cos (c+d x)+a)^3}}{7 a^2}-\frac {(A+C) \tan (c+d x)}{7 d (a \cos (c+d x)+a)^4}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\int \frac {a^2 (52 A+3 C)-6 a^2 (6 A-C) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^2 \left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^2}dx}{5 a^2}-\frac {2 a (6 A-C) \tan (c+d x)}{5 d (a \cos (c+d x)+a)^3}}{7 a^2}-\frac {(A+C) \tan (c+d x)}{7 d (a \cos (c+d x)+a)^4}\) |
| \(\Big \downarrow \) 3457 |
| \(\displaystyle \frac {\frac {\frac {\int \frac {2 \left (a^3 (122 A+3 C)-a^3 (88 A-3 C) \cos (c+d x)\right ) \sec ^2(c+d x)}{\cos (c+d x) a+a}dx}{3 a^2}-\frac {(88 A-3 C) \tan (c+d x)}{3 d (\cos (c+d x)+1)^2}}{5 a^2}-\frac {2 a (6 A-C) \tan (c+d x)}{5 d (a \cos (c+d x)+a)^3}}{7 a^2}-\frac {(A+C) \tan (c+d x)}{7 d (a \cos (c+d x)+a)^4}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {\frac {2 \int \frac {\left (a^3 (122 A+3 C)-a^3 (88 A-3 C) \cos (c+d x)\right ) \sec ^2(c+d x)}{\cos (c+d x) a+a}dx}{3 a^2}-\frac {(88 A-3 C) \tan (c+d x)}{3 d (\cos (c+d x)+1)^2}}{5 a^2}-\frac {2 a (6 A-C) \tan (c+d x)}{5 d (a \cos (c+d x)+a)^3}}{7 a^2}-\frac {(A+C) \tan (c+d x)}{7 d (a \cos (c+d x)+a)^4}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\frac {2 \int \frac {a^3 (122 A+3 C)-a^3 (88 A-3 C) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^2 \left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )}dx}{3 a^2}-\frac {(88 A-3 C) \tan (c+d x)}{3 d (\cos (c+d x)+1)^2}}{5 a^2}-\frac {2 a (6 A-C) \tan (c+d x)}{5 d (a \cos (c+d x)+a)^3}}{7 a^2}-\frac {(A+C) \tan (c+d x)}{7 d (a \cos (c+d x)+a)^4}\) |
| \(\Big \downarrow \) 3457 |
| \(\displaystyle \frac {\frac {\frac {2 \left (\frac {\int \left (a^4 (332 A+3 C)-210 a^4 A \cos (c+d x)\right ) \sec ^2(c+d x)dx}{a^2}-\frac {210 a^3 A \tan (c+d x)}{d (a \cos (c+d x)+a)}\right )}{3 a^2}-\frac {(88 A-3 C) \tan (c+d x)}{3 d (\cos (c+d x)+1)^2}}{5 a^2}-\frac {2 a (6 A-C) \tan (c+d x)}{5 d (a \cos (c+d x)+a)^3}}{7 a^2}-\frac {(A+C) \tan (c+d x)}{7 d (a \cos (c+d x)+a)^4}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\frac {2 \left (\frac {\int \frac {a^4 (332 A+3 C)-210 a^4 A \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^2}dx}{a^2}-\frac {210 a^3 A \tan (c+d x)}{d (a \cos (c+d x)+a)}\right )}{3 a^2}-\frac {(88 A-3 C) \tan (c+d x)}{3 d (\cos (c+d x)+1)^2}}{5 a^2}-\frac {2 a (6 A-C) \tan (c+d x)}{5 d (a \cos (c+d x)+a)^3}}{7 a^2}-\frac {(A+C) \tan (c+d x)}{7 d (a \cos (c+d x)+a)^4}\) |
| \(\Big \downarrow \) 3227 |
| \(\displaystyle \frac {\frac {\frac {2 \left (\frac {a^4 (332 A+3 C) \int \sec ^2(c+d x)dx-210 a^4 A \int \sec (c+d x)dx}{a^2}-\frac {210 a^3 A \tan (c+d x)}{d (a \cos (c+d x)+a)}\right )}{3 a^2}-\frac {(88 A-3 C) \tan (c+d x)}{3 d (\cos (c+d x)+1)^2}}{5 a^2}-\frac {2 a (6 A-C) \tan (c+d x)}{5 d (a \cos (c+d x)+a)^3}}{7 a^2}-\frac {(A+C) \tan (c+d x)}{7 d (a \cos (c+d x)+a)^4}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\frac {2 \left (\frac {a^4 (332 A+3 C) \int \csc \left (c+d x+\frac {\pi }{2}\right )^2dx-210 a^4 A \int \csc \left (c+d x+\frac {\pi }{2}\right )dx}{a^2}-\frac {210 a^3 A \tan (c+d x)}{d (a \cos (c+d x)+a)}\right )}{3 a^2}-\frac {(88 A-3 C) \tan (c+d x)}{3 d (\cos (c+d x)+1)^2}}{5 a^2}-\frac {2 a (6 A-C) \tan (c+d x)}{5 d (a \cos (c+d x)+a)^3}}{7 a^2}-\frac {(A+C) \tan (c+d x)}{7 d (a \cos (c+d x)+a)^4}\) |
| \(\Big \downarrow \) 4254 |
| \(\displaystyle \frac {\frac {\frac {2 \left (\frac {-\frac {a^4 (332 A+3 C) \int 1d(-\tan (c+d x))}{d}-210 a^4 A \int \csc \left (c+d x+\frac {\pi }{2}\right )dx}{a^2}-\frac {210 a^3 A \tan (c+d x)}{d (a \cos (c+d x)+a)}\right )}{3 a^2}-\frac {(88 A-3 C) \tan (c+d x)}{3 d (\cos (c+d x)+1)^2}}{5 a^2}-\frac {2 a (6 A-C) \tan (c+d x)}{5 d (a \cos (c+d x)+a)^3}}{7 a^2}-\frac {(A+C) \tan (c+d x)}{7 d (a \cos (c+d x)+a)^4}\) |
| \(\Big \downarrow \) 24 |
| \(\displaystyle \frac {\frac {\frac {2 \left (\frac {\frac {a^4 (332 A+3 C) \tan (c+d x)}{d}-210 a^4 A \int \csc \left (c+d x+\frac {\pi }{2}\right )dx}{a^2}-\frac {210 a^3 A \tan (c+d x)}{d (a \cos (c+d x)+a)}\right )}{3 a^2}-\frac {(88 A-3 C) \tan (c+d x)}{3 d (\cos (c+d x)+1)^2}}{5 a^2}-\frac {2 a (6 A-C) \tan (c+d x)}{5 d (a \cos (c+d x)+a)^3}}{7 a^2}-\frac {(A+C) \tan (c+d x)}{7 d (a \cos (c+d x)+a)^4}\) |
| \(\Big \downarrow \) 4257 |
| \(\displaystyle \frac {\frac {\frac {2 \left (\frac {\frac {a^4 (332 A+3 C) \tan (c+d x)}{d}-\frac {210 a^4 A \text {arctanh}(\sin (c+d x))}{d}}{a^2}-\frac {210 a^3 A \tan (c+d x)}{d (a \cos (c+d x)+a)}\right )}{3 a^2}-\frac {(88 A-3 C) \tan (c+d x)}{3 d (\cos (c+d x)+1)^2}}{5 a^2}-\frac {2 a (6 A-C) \tan (c+d x)}{5 d (a \cos (c+d x)+a)^3}}{7 a^2}-\frac {(A+C) \tan (c+d x)}{7 d (a \cos (c+d x)+a)^4}\) |
Input:
Int[((A + C*Cos[c + d*x]^2)*Sec[c + d*x]^2)/(a + a*Cos[c + d*x])^4,x]
Output:
-1/7*((A + C)*Tan[c + d*x])/(d*(a + a*Cos[c + d*x])^4) + ((-2*a*(6*A - C)* Tan[c + d*x])/(5*d*(a + a*Cos[c + d*x])^3) + (-1/3*((88*A - 3*C)*Tan[c + d *x])/(d*(1 + Cos[c + d*x])^2) + (2*((-210*a^3*A*Tan[c + d*x])/(d*(a + a*Co s[c + d*x])) + ((-210*a^4*A*ArcTanh[Sin[c + d*x]])/d + (a^4*(332*A + 3*C)* Tan[c + d*x])/d)/a^2))/(3*a^2))/(5*a^2))/(7*a^2)
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x _)]), x_Symbol] :> Simp[c Int[(b*Sin[e + f*x])^m, x], x] + Simp[d/b Int [(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x]
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim p[b*(A*b - a*B)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x])^( n + 1)/(a*f*(2*m + 1)*(b*c - a*d))), x] + Simp[1/(a*(2*m + 1)*(b*c - a*d)) Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[B*(a*c*m + b *d*(n + 1)) + A*(b*c*(m + 1) - a*d*(2*m + n + 2)) + d*(A*b - a*B)*(m + n + 2)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, x] && NeQ [b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[m, -2^(-1)] && !GtQ[n, 0] && IntegerQ[2*m] && (IntegerQ[2*n] || EqQ[c, 0])
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((A_.) + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[a*(A + C)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x])^(n + 1)/(f*(b*c - a*d)*(2*m + 1))), x] + Simp[1/(b*(b*c - a*d)*(2*m + 1)) I nt[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[A*(a*c*(m + 1) - b*d*(2*m + n + 2)) - C*(a*c*m + b*d*(n + 1)) + (a*A*d*(m + n + 2) + C*(b* c*(2*m + 1) - a*d*(m - n - 1)))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, C, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[m, -2^(-1)]
Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> Simp[-d^(-1) Subst[Int[Exp andIntegrand[(1 + x^2)^(n/2 - 1), x], x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]
Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]
Time = 0.50 (sec) , antiderivative size = 155, normalized size of antiderivative = 0.96
| method | result | size |
| parallelrisch | \(\frac {3360 \cos \left (d x +c \right ) A \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )-3360 \cos \left (d x +c \right ) A \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+83 \left (\frac {15 \left (110 A +\frac {3 C}{2}\right ) \cos \left (2 d x +2 c \right )}{83}+\frac {\left (559 A +6 C \right ) \cos \left (3 d x +3 c \right )}{83}+\left (A +\frac {3 C}{332}\right ) \cos \left (4 d x +4 c \right )+\frac {\left (2861 A +54 C \right ) \cos \left (d x +c \right )}{83}+\frac {1672 A}{83}+\frac {87 C}{332}\right ) \sec \left (\frac {d x}{2}+\frac {c}{2}\right )^{6} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{840 d \,a^{4} \cos \left (d x +c \right )}\) | \(155\) |
| derivativedivides | \(\frac {\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7} A}{7}+\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7} C}{7}+\frac {7 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} A}{5}+\frac {3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} C}{5}+\frac {23 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} A}{3}+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} C +49 A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+C \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-\frac {8 A}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1}-32 A \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-\frac {8 A}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1}+32 A \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{8 d \,a^{4}}\) | \(178\) |
| default | \(\frac {\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7} A}{7}+\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7} C}{7}+\frac {7 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} A}{5}+\frac {3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} C}{5}+\frac {23 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} A}{3}+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} C +49 A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+C \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-\frac {8 A}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1}-32 A \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-\frac {8 A}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1}+32 A \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{8 d \,a^{4}}\) | \(178\) |
| risch | \(\frac {4 i \left (210 A \,{\mathrm e}^{8 i \left (d x +c \right )}+1470 A \,{\mathrm e}^{7 i \left (d x +c \right )}+4550 A \,{\mathrm e}^{6 i \left (d x +c \right )}+8330 A \,{\mathrm e}^{5 i \left (d x +c \right )}+105 C \,{\mathrm e}^{5 i \left (d x +c \right )}+10262 A \,{\mathrm e}^{4 i \left (d x +c \right )}+63 C \,{\mathrm e}^{4 i \left (d x +c \right )}+9394 A \,{\mathrm e}^{3 i \left (d x +c \right )}+126 C \,{\mathrm e}^{3 i \left (d x +c \right )}+5834 A \,{\mathrm e}^{2 i \left (d x +c \right )}+66 C \,{\mathrm e}^{2 i \left (d x +c \right )}+2114 A \,{\mathrm e}^{i \left (d x +c \right )}+21 C \,{\mathrm e}^{i \left (d x +c \right )}+332 A +3 C \right )}{105 d \,a^{4} \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )^{7} \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}+\frac {4 A \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{a^{4} d}-\frac {4 A \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{a^{4} d}\) | \(244\) |
| norman | \(\frac {\frac {\left (A +C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{13}}{56 a d}+\frac {\left (27 A +13 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{140 a d}-\frac {\left (65 A +C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{8 a d}-\frac {\left (133 A +3 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{12 a d}+\frac {\left (359 A -9 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{120 a d}+\frac {\left (937 A +153 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{840 a d}+\frac {\left (1447 A +33 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{210 a d}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right ) a^{3}}+\frac {4 A \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{a^{4} d}-\frac {4 A \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{a^{4} d}\) | \(253\) |
Input:
int((A+C*cos(d*x+c)^2)*sec(d*x+c)^2/(a+a*cos(d*x+c))^4,x,method=_RETURNVER BOSE)
Output:
1/840*(3360*cos(d*x+c)*A*ln(tan(1/2*d*x+1/2*c)-1)-3360*cos(d*x+c)*A*ln(tan (1/2*d*x+1/2*c)+1)+83*(15/83*(110*A+3/2*C)*cos(2*d*x+2*c)+1/83*(559*A+6*C) *cos(3*d*x+3*c)+(A+3/332*C)*cos(4*d*x+4*c)+1/83*(2861*A+54*C)*cos(d*x+c)+1 672/83*A+87/332*C)*sec(1/2*d*x+1/2*c)^6*tan(1/2*d*x+1/2*c))/d/a^4/cos(d*x+ c)
Time = 0.12 (sec) , antiderivative size = 277, normalized size of antiderivative = 1.72 \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^2(c+d x)}{(a+a \cos (c+d x))^4} \, dx=-\frac {210 \, {\left (A \cos \left (d x + c\right )^{5} + 4 \, A \cos \left (d x + c\right )^{4} + 6 \, A \cos \left (d x + c\right )^{3} + 4 \, A \cos \left (d x + c\right )^{2} + A \cos \left (d x + c\right )\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 210 \, {\left (A \cos \left (d x + c\right )^{5} + 4 \, A \cos \left (d x + c\right )^{4} + 6 \, A \cos \left (d x + c\right )^{3} + 4 \, A \cos \left (d x + c\right )^{2} + A \cos \left (d x + c\right )\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - {\left (2 \, {\left (332 \, A + 3 \, C\right )} \cos \left (d x + c\right )^{4} + 4 \, {\left (559 \, A + 6 \, C\right )} \cos \left (d x + c\right )^{3} + {\left (2636 \, A + 39 \, C\right )} \cos \left (d x + c\right )^{2} + 4 \, {\left (296 \, A + 9 \, C\right )} \cos \left (d x + c\right ) + 105 \, A\right )} \sin \left (d x + c\right )}{105 \, {\left (a^{4} d \cos \left (d x + c\right )^{5} + 4 \, a^{4} d \cos \left (d x + c\right )^{4} + 6 \, a^{4} d \cos \left (d x + c\right )^{3} + 4 \, a^{4} d \cos \left (d x + c\right )^{2} + a^{4} d \cos \left (d x + c\right )\right )}} \] Input:
integrate((A+C*cos(d*x+c)^2)*sec(d*x+c)^2/(a+a*cos(d*x+c))^4,x, algorithm= "fricas")
Output:
-1/105*(210*(A*cos(d*x + c)^5 + 4*A*cos(d*x + c)^4 + 6*A*cos(d*x + c)^3 + 4*A*cos(d*x + c)^2 + A*cos(d*x + c))*log(sin(d*x + c) + 1) - 210*(A*cos(d* x + c)^5 + 4*A*cos(d*x + c)^4 + 6*A*cos(d*x + c)^3 + 4*A*cos(d*x + c)^2 + A*cos(d*x + c))*log(-sin(d*x + c) + 1) - (2*(332*A + 3*C)*cos(d*x + c)^4 + 4*(559*A + 6*C)*cos(d*x + c)^3 + (2636*A + 39*C)*cos(d*x + c)^2 + 4*(296* A + 9*C)*cos(d*x + c) + 105*A)*sin(d*x + c))/(a^4*d*cos(d*x + c)^5 + 4*a^4 *d*cos(d*x + c)^4 + 6*a^4*d*cos(d*x + c)^3 + 4*a^4*d*cos(d*x + c)^2 + a^4* d*cos(d*x + c))
Timed out. \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^2(c+d x)}{(a+a \cos (c+d x))^4} \, dx=\text {Timed out} \] Input:
integrate((A+C*cos(d*x+c)**2)*sec(d*x+c)**2/(a+a*cos(d*x+c))**4,x)
Output:
Timed out
Time = 0.05 (sec) , antiderivative size = 274, normalized size of antiderivative = 1.70 \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^2(c+d x)}{(a+a \cos (c+d x))^4} \, dx=\frac {A {\left (\frac {1680 \, \sin \left (d x + c\right )}{{\left (a^{4} - \frac {a^{4} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}\right )} {\left (\cos \left (d x + c\right ) + 1\right )}} + \frac {\frac {5145 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {805 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {147 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + \frac {15 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}}}{a^{4}} - \frac {3360 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a^{4}} + \frac {3360 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a^{4}}\right )} + \frac {3 \, C {\left (\frac {35 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {35 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {21 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + \frac {5 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}}\right )}}{a^{4}}}{840 \, d} \] Input:
integrate((A+C*cos(d*x+c)^2)*sec(d*x+c)^2/(a+a*cos(d*x+c))^4,x, algorithm= "maxima")
Output:
1/840*(A*(1680*sin(d*x + c)/((a^4 - a^4*sin(d*x + c)^2/(cos(d*x + c) + 1)^ 2)*(cos(d*x + c) + 1)) + (5145*sin(d*x + c)/(cos(d*x + c) + 1) + 805*sin(d *x + c)^3/(cos(d*x + c) + 1)^3 + 147*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 + 15*sin(d*x + c)^7/(cos(d*x + c) + 1)^7)/a^4 - 3360*log(sin(d*x + c)/(cos( d*x + c) + 1) + 1)/a^4 + 3360*log(sin(d*x + c)/(cos(d*x + c) + 1) - 1)/a^4 ) + 3*C*(35*sin(d*x + c)/(cos(d*x + c) + 1) + 35*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 21*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 + 5*sin(d*x + c)^7/(co s(d*x + c) + 1)^7)/a^4)/d
Time = 0.34 (sec) , antiderivative size = 212, normalized size of antiderivative = 1.32 \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^2(c+d x)}{(a+a \cos (c+d x))^4} \, dx=-\frac {\frac {3360 \, A \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{a^{4}} - \frac {3360 \, A \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{a^{4}} + \frac {1680 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )} a^{4}} - \frac {15 \, A a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 15 \, C a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 147 \, A a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 63 \, C a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 805 \, A a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 105 \, C a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 5145 \, A a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 105 \, C a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{28}}}{840 \, d} \] Input:
integrate((A+C*cos(d*x+c)^2)*sec(d*x+c)^2/(a+a*cos(d*x+c))^4,x, algorithm= "giac")
Output:
-1/840*(3360*A*log(abs(tan(1/2*d*x + 1/2*c) + 1))/a^4 - 3360*A*log(abs(tan (1/2*d*x + 1/2*c) - 1))/a^4 + 1680*A*tan(1/2*d*x + 1/2*c)/((tan(1/2*d*x + 1/2*c)^2 - 1)*a^4) - (15*A*a^24*tan(1/2*d*x + 1/2*c)^7 + 15*C*a^24*tan(1/2 *d*x + 1/2*c)^7 + 147*A*a^24*tan(1/2*d*x + 1/2*c)^5 + 63*C*a^24*tan(1/2*d* x + 1/2*c)^5 + 805*A*a^24*tan(1/2*d*x + 1/2*c)^3 + 105*C*a^24*tan(1/2*d*x + 1/2*c)^3 + 5145*A*a^24*tan(1/2*d*x + 1/2*c) + 105*C*a^24*tan(1/2*d*x + 1 /2*c))/a^28)/d
Time = 0.15 (sec) , antiderivative size = 204, normalized size of antiderivative = 1.27 \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^2(c+d x)}{(a+a \cos (c+d x))^4} \, dx=\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (\frac {A+C}{20\,a^4}+\frac {5\,A+C}{40\,a^4}\right )}{d}+\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {A+C}{2\,a^4}+\frac {3\,\left (5\,A+C\right )}{8\,a^4}+\frac {3\,\left (10\,A-2\,C\right )}{8\,a^4}\right )}{d}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (\frac {A+C}{8\,a^4}+\frac {5\,A+C}{12\,a^4}+\frac {10\,A-2\,C}{24\,a^4}\right )}{d}-\frac {8\,A\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{a^4\,d}-\frac {2\,A\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left (a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-a^4\right )}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7\,\left (A+C\right )}{56\,a^4\,d} \] Input:
int((A + C*cos(c + d*x)^2)/(cos(c + d*x)^2*(a + a*cos(c + d*x))^4),x)
Output:
(tan(c/2 + (d*x)/2)^5*((A + C)/(20*a^4) + (5*A + C)/(40*a^4)))/d + (tan(c/ 2 + (d*x)/2)*((A + C)/(2*a^4) + (3*(5*A + C))/(8*a^4) + (3*(10*A - 2*C))/( 8*a^4)))/d + (tan(c/2 + (d*x)/2)^3*((A + C)/(8*a^4) + (5*A + C)/(12*a^4) + (10*A - 2*C)/(24*a^4)))/d - (8*A*atanh(tan(c/2 + (d*x)/2)))/(a^4*d) - (2* A*tan(c/2 + (d*x)/2))/(d*(a^4*tan(c/2 + (d*x)/2)^2 - a^4)) + (tan(c/2 + (d *x)/2)^7*(A + C))/(56*a^4*d)
Time = 0.18 (sec) , antiderivative size = 228, normalized size of antiderivative = 1.42 \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^2(c+d x)}{(a+a \cos (c+d x))^4} \, dx=\frac {3360 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -3360 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) a -3360 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a +3360 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) a +15 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9} a +15 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9} c +132 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7} a +48 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7} c +658 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} a +42 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} c +4340 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} a -6825 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) a -105 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) c}{840 a^{4} d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )} \] Input:
int((A+C*cos(d*x+c)^2)*sec(d*x+c)^2/(a+a*cos(d*x+c))^4,x)
Output:
(3360*log(tan((c + d*x)/2) - 1)*tan((c + d*x)/2)**2*a - 3360*log(tan((c + d*x)/2) - 1)*a - 3360*log(tan((c + d*x)/2) + 1)*tan((c + d*x)/2)**2*a + 33 60*log(tan((c + d*x)/2) + 1)*a + 15*tan((c + d*x)/2)**9*a + 15*tan((c + d* x)/2)**9*c + 132*tan((c + d*x)/2)**7*a + 48*tan((c + d*x)/2)**7*c + 658*ta n((c + d*x)/2)**5*a + 42*tan((c + d*x)/2)**5*c + 4340*tan((c + d*x)/2)**3* a - 6825*tan((c + d*x)/2)*a - 105*tan((c + d*x)/2)*c)/(840*a**4*d*(tan((c + d*x)/2)**2 - 1))